Denominator Summation Version 2

$\dfrac{1}{a+b} = \dfrac{1}{a} + \dfrac{1}{b}\\ \dfrac{1}{a+b} = \dfrac{a+b}{ab}\\ (a+b)^2 = ab\\ a^2+2ab+b^2 - ab = 0\\ a^2 + ab + b^2 = 0$

Now, we let $b$ be an integer constant, and $a$ be the variable. This gives us a quadratic equation. The discriminant,

$\Delta =b^2 - 4(1)(b^2) =-3b^2$

Since we know that $b^2 \geq 0$ for all values of $b$

$b^2 \geq 0\\ -3b^2 \leq 0\\ \Delta \leq 0$

This implies that for $a$ to have a real value,

$\Delta = 0 \implies -3b^2 = 0 \implies b=0$

However, if $b=0$ , $\dfrac{1}{b}$ is undefined.

Therefore, no ordered pairs of integers $(a,b)$ satisfy this equation. The answer is $\boxed{0}$

Note: There are actually no real values $a$ and $b$ that satisfy this equation either

$\dfrac{1}{a+b} = \dfrac{1}{a}+\dfrac{1}{b}$

$\therefore ab=(a+b)^2 \implies ab\geq 0$

On solving,

$a^2+b^2+2ab=ab$ $\therefore a^2+b^2=-ab \implies ab\leq 0$

But $ab\geq 0$ . So any one of $(a,b)$ must be zero which is not possible as inverse of any of two is not possible. So it is not possible. So there exists no such $(a,b)$

Best way is to plot a graph.For the equation. $a^2-ab+b^2=0$ .

Of course $a \not = 0, b \not = 0$ so consider when they are the same sign. If $a,b$ are both positive then $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{a+b} < \dfrac{1}{a} \implies \dfrac{1}{b} < 0$ which is a contradiction. If $a,b$ are both negative then $(-a,-b)$ will also be a solution which as we shown can't happen so if a solution exists, $a,b$ must have opposite signs.

Instead let's look at $\dfrac{1}{a-b} = \dfrac{1}{a} - \dfrac{1}{b} = \dfrac{a-b}{ab}$ which is the original question with $-b$ instead of $b$ such that $a,b$ are both positive. We can't have $a=b$ because $a > 0 > b$ so we can multiply this equation by $ab(a-b$ to get $ab = (a-b)^2 \implies 3ab = a^2 + b^2$

Now consider $a^2 + b^2 \equiv 0 \pmod{3}$ . By considering the quadratic residue modulo 3. We must have $a \equiv b \equiv 0 \pmod 3$ . Let $a=3a', b=3b'$ for some integers $a',b'$ but now this means $\dfrac{1}{3a' - 3b'} = \dfrac{1}{3a'} - \dfrac{1}{3b'} \implies \dfrac{1}{a'-b'} = \dfrac{1}{a'} - \dfrac{1}{b'}$ and so by Fermats proof by infinite decent, there must exist no solutions to this equation. Again if both $a,b$ are negative then taking the negation would imply a solution where $a,b$ are both positive which is again a contradiction.

Therefore there are no solution to $\dfrac{1}{a+b} = \dfrac{1}{a} + \dfrac{1}{b}$